## Functions of the form \(y=a{b}^{(x+p)}+q\)

Contents

We now consider exponential functions of the form \(y=a{b}^{(x+p)}+q\) and the effects of parameter \(p\).

## Optional Investigation: The effects of \(a\), \(p\) and \(q\) on an exponential graph

On the same system of axes, plot the following graphs:

- \(y_1 = 2^x\)
- \(y_2 = 2^{(x – 2)}\)
- \(y_3 = 2^{(x – 1)}\)
- \(y_4 = 2^{(x + 1)}\)
- \(y_5 = 2^{(x + 2)}\)

Use your sketches of the functions above to complete the following table:

**\(y_1\)****\(y_2\)****\(y_3\)****\(y_4\)****\(y_5\)****intercept(s)****asymptote****domain****range****effect of \(p\)**On the same system of axes, plot the following graphs:

- \(y_1 = 2^{(x – 1)} + 2\)
- \(y_2 = 3 \times 2^{(x – 1)} + 2\)
- \(y_3 = \cfrac{1}{2} \times 2^{(x – 1)} + 2\)
- \(y_4 = 0 \times 2^{(x – 1)} + 2\)
- \(y_5 = -3 \times 2^{(x – 1)} + 2\)

Use your sketches of the functions above to complete the following table:

**\(y_1\)****\(y_2\)****\(y_3\)****\(y_4\)****\(y_5\)****intercept(s)****asymptotes****domain****range****effect of \(a\)**

**The effect of the parameters on \(y = ab^{x + p} + q\)**

The effect of \(p\) is a horizontal shift because all points are moved the same distance in the same direction (the entire graph slides to the left or to the right).

For \(p>0\), the graph is shifted to the left by \(p\) units.

For \(p<0\), the graph is shifted to the right by \(p\) units.

The effect of \(q\) is a vertical shift. The value of \(q\) also affects the horizontal asymptotes, the line \(y = q\).

The value of \(a\) affects the shape of the graph and its position relative to the horizontal asymptote.

For \(a>0\), the graph lies above the horizontal asymptote, \(y = q\).

For \(a<0\), the graph lies below the horizontal asymptote, \(y = q\).

\(p>0\) | \(p<0\) | |||

\(a<0\) | \(a>0\) | \(a<0\) | \(a>0\) | |

\(q>0\) | ||||

\(q<0\) |

### Discovering the characteristics

For functions of the general form: \(f(x) = y = ab^{(x+p)} + q\):

**Domain and range**

The domain is \(\{x:x\in ℝ\}\) because there is no value of \(x\) for which \(f(x)\) is undefined.

The range of \(f(x)\) depends on whether the value for \(a\) is positive or negative.

If \(a>0\) we have: \begin{align*} {b}^{(x+p)} & > 0 \\ a {b}^{(x+p)} & > 0 \\ a {b}^{(x+p)} + q & > q \\ f(x) & > q\end{align*} The range is therefore \(\{ y: y > q, y \in \mathbb{R} \}\).

Similarly, if \(a < 0\), the range is \(\{ y: y < q, y \in \mathbb{R} \}\).

## Example

### Question

State the domain and range for \(g(x) = 5 \times 3^{(x+1)} – 1\).

### Determine the domain

The domain is \(\{x: x \in \mathbb{R} \}\) because there is no value of \(x\) for which \(g(x)\) is undefined.

### Determine the range

The range of \(g(x)\) can be calculated from: \begin{align*} 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} & > 0\\ 5 \times 3^{(x+1)} – 1 & > -1\\ \therefore g(x) & > -1 \end{align*} Therefore the range is \(\{g(x): g(x) > -1 \}\) or in interval notation \((-1; \infty)\).

**Intercepts**

**The \(y\)-intercept:**

To calculate the \(y\)-intercept we let \(x=0\). For example, the \(y\)-intercept of \(g(x) = 3 \times 2^{(x + 1)} + 2\) is determined by setting \(x=0\): \begin{align*} g(0) &= 3 \times 2^{(0 + 1)} + 2 \\ &= 3 \times 2 + 2\\ &= 8 \end{align*} This gives the point \((0;8)\).

**The \(x\)-intercept:**

To calculate the \(x\)-intercept we let \(y=0\). For example, the \(x\)-intercept of \(g(x) = 3 \times 2^{(x + 1)} + 2\) is determined by setting \(y=0\): \begin{align*} 0 &= 3 \times 2^{(x + 1)} + 2 \\ -2 &= 3 \times 2^{(x + 1)} \\ -\cfrac{2}{3} &= 2^{(x + 1)} \end{align*} which has no real solutions. Therefore, the graph of \(g(x)\) lies above the \(x\)-axis and does not have any \(x\)-intercepts.

**Asymptote**

Exponential functions of the form \(y = ab^{(x+p)} + q\) have a horizontal asymptote, the line \(y = q\).

## Example

### Question

Determine the asymptote for \(y = 5 \times 3^{(x+1)} – 1\).

### Determine the asymptote

The asymptote of \(g(x)\) can be calculated as: \begin{align*} 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} & \ne 0\\ 5 \times 3^{(x+1)} – 1 & \ne -1\\ \therefore y & \ne -1 \end{align*} Therefore the asymptote is the line \(y = -1\).

### Sketching graphs of the form \(f(x)=a{b}^{(x+p)}+q\)

In order to sketch graphs of functions of the form, \(f(x)=a{b}^{(x+p)}+q\), we need to determine five characteristics:

shape

\(y\)-intercept

\(x\)-intercept

asymptote

domain and range

## Example

### Question

Sketch the graph of \(2y = \text{10} \times 2^{(x+1)} – 5\).

Mark the intercept(s) and asymptote. State the domain and range of the function.

### Examine the equation of the form \(y = ab^{(x + p)} + q\)

We notice that \(a > 0\) and \(b > 1\), therefore the function is increasing.

### Determine the \(y\)-intercept

The \(y\)-intercept is obtained by letting \(x = 0\): \begin{align*} 2y &= \text{10} \times 2^{(0+1)} – 5\\ &= \text{10} \times 2 – 5\\ &= \text{15}\\ \therefore y &= 7\cfrac{1}{2} \end{align*} This gives the point \((0;7\cfrac{1}{2})\).

### Determine the \(x\)-intercept

The \(x\)-intercept is obtained by letting \(y = 0\): \begin{align*} 0 &= \text{10} \times 2^{(x+1)} – 5\\ 5 &= \text{10} \times 2^{(x+1)} \\ \cfrac{1}{2} &= 2^{(x+1)}\\ 2^{-1} &= 2^{(x+1)}\\ \therefore -1 &= x + 1 \quad \text{(same base)}\\ -2 &= x \end{align*} This gives the point \((-2;0)\).

### Determine the asymptote

The horizontal asymptote is the line \(y = -\cfrac{5}{2}\).

### Plot the points and sketch the graph

### State the domain and range

Domain: \(\{ x: x \in \mathbb{R} \}\)

Range: \(\{ y: y > -\cfrac{5}{2}, y \in \mathbb{R} \}\)

**Finding the equation of an exponential function from the graph**

## Example

### Question

Use the given graph of \(y = -2 \times 3^{(x + p)} + q\) to determine the values of \(p\) and \(q\).

### Examine the equation of the form \(y = ab^{(x + p)} + q\)

From the graph we see that the function is decreasing. We also note that \(a = -2\) and \(b = 3\).

We need to solve for \(p\) and \(q\).

### Use the asymptote to determine \(q\)

The horizontal asymptote \(y = 6\) is given, therefore we know that \(q = 6\). \[y = -2 \times 3^{(x + p)} + 6\]

### Use the \(x\)-intercept to determine \(p\)

Substitute \((2;0)\) into the equation and solve for \(p\): \begin{align*} y &= -2 \times 3^{(x + p)} + 6 \\ 0 &= -2 \times 3^{(2 + p)} + 6 \\ -6 &= -2 \times 3^{(2 + p)} \\ 3 &= 3^{(2 + p)} \\ \therefore 1 &= 2 + p \quad \text{(same base)}\\ \therefore p &= -1 \end{align*}

### Write the final answer

\[y = -2 \times 3^{(x – 1)} + 6\]